= Solution
For a <plane wave> with <amplitude> $\alpha$, <wavenumber> $k>0$ and <angular frequency> $\omega>0$, the <dispersion relation> $\omega=\omega(k)$ specifies which frequencies the governing equations permit. A point of fixed <wave phase> obeys $kx-\omega t=\text{constant}$, so the <phase speed> is
$$
\boxed{c_p=\frac\omega k}.
$$
It is the speed of individual crests. For a narrow <wave packet>, two nearby <plane waves> with differences $\Delta k$ and $\Delta\omega$ have a slowly varying <wave envelope> depending on $\Delta k\,x-\Delta\omega\,t$. Taking the limit gives the <group velocity>
$$
\boxed{c_g=\frac{d\omega}{dk}}.
$$
This is the envelope's propagation speed and, in the present linear lossless wave model, the <energy> transport speed for a narrow packet.
A <dispersive wave> has <phase speed> dependent on <wavenumber>, so different components can change their relative phases and alter the packet's shape. A <nondispersive wave> on the chosen propagation branch has constant <phase speed> over the relevant band:
$$
\boxed{\omega(k)=ck,\qquad c_p=c_g=c}.
$$
Indeed $dc_p/dk=(c_g-c_p)/k$, so equality throughout a band is equivalent to constant <phase speed> there. Equality at an isolated <wavenumber> does not make the entire wave family nondispersive.
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